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(D) Given the curve equation is $y = \cos(x + y)$.Differentiating both sides with respect to $x$,we get:$\frac{dy}{dx} = -\sin(x + y) \left(1 + \frac{

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$\frac{\pi}{2}$

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(D) Given the curve equation is $y = \cos(x + y)$.Differentiating both sides with respect to $x$,we get:$\frac{dy}{dx} = -\sin(x + y) \left(1 + \frac{dy}{dx}\right) \dots (1)$The equation of the tangent is $x + 2y = k$,which can be written as $y = -\frac{1}{2}x + \frac{k}{2}$.The slope of the tangent is $m = -\frac{1}{2}$.Substituting $\frac{dy}{dx} = -\frac{1}{2}$ into equation $(1)$:$-\frac{1}{2} = -\sin(x + y) \left(1 - \frac{1}{2}\right)$$-\frac{1}{2} = -\sin(x + y) \left(\frac{1}{2}\right)$$\sin(x + y) = 1$This implies $x + y = \frac{\pi}{2}$.Substituting $x + y = \frac{\pi}{2}$ into the original curve equation $y = \cos(x + y)$:$y = \cos\left(\frac{\pi}{2}\right) = 0$.Since $y = 0$ and $x + y = \frac{\pi}{2}$,we have $x = \frac{\pi}{2}$.Now,substitute the point $(\frac{\pi}{2}, 0)$ into the tangent equation $x + 2y = k$:$\frac{\pi}{2} + 2(0) = k$$k = \frac{\pi}{2}$.

If an equation of a tangent to the curve $y = \cos(x + y)$,where $-1 - \pi \le x \le 1 + \pi$,is $x + 2y = k$,then $k$ is equal to

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